def migratoryBirds(arr) freq = {}

# Step 1: Build frequency hash manually
for i in arr
  if freq.has_key?(i)
    freq[i] += 1
  else
    freq[i] = 1
  end
end

# Step 2: Find the highest frequency
max_freq = 0
for key in freq.keys
  if freq[key] > max_freq
    max_freq = freq[key]
  end
end

# Step 3: Among all bird types with max frequency, find the smallest ID
min_id = nil
for key in freq.keys
  if freq[key] == max_freq
    if min_id.nil? || key < min_id
      min_id = key
    end
  end
end

return min_id

end

Python

def migratoryBirds(arr): # Write your code here res = [] for i in range(5): res.append(arr.count(i+1)) return res.index(max(res))+1

Why can't I use statistics.mode(arr) to solve it.

public static int migratoryBirds(List<Integer> arr) {
    Comparator<Long> longComp = Long::compare;
    Map<Integer, Long> countMap = arr.stream().collect(Collectors.groupingBy((Integer i)->i, Collectors.counting()));
    long maxFrequency = countMap.entrySet().stream().sorted(Map.Entry.comparingByValue(longComp.reversed())).
    findFirst().get().getValue();
    Integer key = countMap.entrySet().stream().filter(en -> en.getValue() == maxFrequency).sorted(Map.Entry.comparingByKey()).findFirst().get().getKey();
    return key;
}   

Here is a python solution with O(n + k) time and O(k) space:

def migratoryBirds(arr):
    count = {}
    for n in arr:
        if n not in count:
            count[n] = 0
        count[n] += 1
        
    maxval = max(count.values())
    
    best = []
    for key, val in count.items():
        if val == maxval:
            best.append(key)             

    return min(best)